Ca + 2HCl = CaCl₂ + H₂
m(Ca)=60 g
c=1.25 mol/l
M(Ca)=40g/mol
v-?
m(Ca)/M(Ca)=m(HCl)/[2M(HCl)]=n(HCl)/2
n(HCl)=2m(Ca)/M(Ca)
n(HCl)=cv
cv=2m(Ca)/M(Ca)
v=2m(Ca)/{cM(Ca)}
v=2·60g/mol/{1.25mol/l·40g/mol}=2.4 l = 2400 ml
2400 milliliters of a 1.25 molar HCl solution would be needed
Answer:
1.24 × 10³ kPa
Explanation:
Step 1: Given data
- Initial pressure of the gas (P₁): 34.5 kPa
- Initial volume of the can (V₁): 473 mL
- Final pressure of the gas (P₂): ?
- Final volume of the can (V₂): 13.16 mL
Step 2: Calculate the final pressure of the gas in the can
If we assume that the gas in the can behaves as an ideal gas and that the temperature remains constant, we can calculate the final pressure of the gas using Boyle's law.
P₁ × V₁ = P₂ × V₂
P₂ = P₁ × V₁ / V₂
P₂ = 34.5 kPa × 473 mL / 13.16 mL = 1.24 × 10³ kPa
Explanation:
The substance that is in excess that doesn't get used up as a reactant is called limiting reactant.
The answer I would choose is the third one
